Problem: $\lim_{x\to\infty}\dfrac{2x^2+5}{6x^2-9}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $\dfrac{1}{9}$ (Choice C) C $\dfrac{1}{3}$ (Choice D) D $\infty$
Answer: $\lim_{x\to\infty} 2x^2+5=\infty$ and $\lim_{x\to\infty} 6x^2-9=\infty$, so $\lim_{x\to\infty}\dfrac{2x^2+5}{6x^2-9}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{2x^2+5}{6x^2-9} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[2x^2+5\right]}{\dfrac{d}{dx}[6x^2-9]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{4x}{12x} \\\\ &=\lim_{x\to\infty}\dfrac{1}{3} \\\\ &=\dfrac{1}{3} \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[2x^2+5\right]}{\dfrac{d}{dx}[6x^2-9]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{2x^2+5}{6x^2-9}=\dfrac{1}{3}$.